Local Semicircle Law Under Fourth Moment Condition

Abstract

We consider a random symmetric matrix $\mathbf{X}= [X_{jk}]_{j,k=1}^n$ with upper triangular entries being independent random variables with mean zero and unit variance. Assuming that $\max _{jk} {{\,\mathrm{\mathbb {E}}\,}}|X_{jk}|^{4+\delta } < \infty , \delta > 0$, it was proved in Götze et al. (Bernoulli 24(3):2358–2400, 2018) that with high probability the typical distance between the Stieltjes transforms $m_n(z)$, $z = u + i v$, of the empirical spectral distribution (ESD) and the Stieltjes transforms $m_{\text {sc}}(z)$ of the semicircle law is of order $(nv)^{-1} \log n$. The aim of this paper is to remove $\delta >0$ and show that this result still holds if we assume that $\max _{jk} {{\,\mathrm{\mathbb {E}}\,}}|X_{jk}|^{4} < \infty$. We also discuss applications to the rate of convergence of the ESD to the semicircle law in the Kolmogorov distance, rates of localization of the eigenvalues around the classical positions and rates of delocalization of eigenvectors.

Auxiliary Results

1.1 Truncation

In this section, we will show that conditions $\mathbf{(C0)}$ allows us to assume that for all $1 \le j,k \le n$ we have $|X_{jk}| \le \sqrt{n}/\overline{R}$, where $\overline{R}$ is some positive constant .

Let $\hat{X}_{jk}: = X_{jk} {{\,\mathrm{\mathbb {1}}\,}}[|X_{jk}| \le \sqrt{n}/\overline{R}]$, $\tilde{X}_{jk}: = X_{jk} {{\,\mathrm{\mathbb {1}}\,}}[|X_{jk}| \ge \sqrt{n}/\overline{R}] - {{\,\mathrm{\mathbb {E}}\,}}X_{jk} {{\,\mathrm{\mathbb {1}}\,}}[|X_{jk}| \ge \sqrt{n}/\overline{R} ]$ and finally ${\breve{X}}_{jk}: = \tilde{X}_{jk} \sigma ^{-1}$, where $\sigma ^2: = {{\,\mathrm{\mathbb {E}}\,}}|\tilde{X}_{11}|^2$. We denote symmetric random matrices by $\hat{\mathbf{X}}, \tilde{\mathbf{X}}$ and ${\breve{\mathbf{X}}}$ formed from $\hat{X}_{jk}, \tilde{X}_{jk}$ and ${\breve{X}}_{jk}$, respectively. Similar notations are used for the corresponding resolvent matrices, ESD and Stieltjes transforms.

Lemma A.1

Assuming the conditions $\mathbf{(C0)}$ we have for all $1 \le p \le A_1 \log n$

$$\begin{aligned} {{\,\mathrm{\mathbb {E}}\,}}|m_n(z) - \hat{m}_n(z)|^p \le \left( \frac{C \overline{R}^4}{nv}\right) ^p. \end{aligned}$$

Proof

From Bai’s rank inequality (see [3, Theorem A.43]) we conclude that

$$\begin{aligned} \sup _{x \in \mathbb {R}} |\mu _n((-\infty , x]) - \hat{\mu }_n((-\infty , x])| \le \frac{1}{n} {{\,\mathrm{Rank}\,}}(\mathbf{X}- \hat{\mathbf{X}}) \le \frac{1}{n} \sum _{j,k=1}^n {{\,\mathrm{\mathbb {1}}\,}}[|X_{jk}| \ge \sqrt{n}/\overline{R}]. \end{aligned}$$

Integrating by parts we get

$$\begin{aligned} {{\,\mathrm{\mathbb {E}}\,}}|m_n(z) - \hat{m}_n(z)|^p \le \frac{1}{(nv)^p} {{\,\mathrm{\mathbb {E}}\,}}\left( \sum _{j,k=1}^n {{\,\mathrm{\mathbb {1}}\,}}[|X_{jk}| \ge \sqrt{n}/\overline{R}] \right) ^p. \end{aligned}$$

It is easy to see that

$$\begin{aligned} \left( \sum _{j,k=1}^n {{\,\mathrm{\mathbb {E}}\,}}{{\,\mathrm{\mathbb {1}}\,}}[|X_{jk}| \ge \sqrt{n} /\overline{R}] \right) ^p \le C^p \overline{R}^{4p}. \end{aligned}$$

Applying Rosenthal’s inequality, [21], we get that

$$\begin{aligned}&{{\,\mathrm{\mathbb {E}}\,}}\left( \sum _{j,k=1}^n [{{\,\mathrm{\mathbb {1}}\,}}[|X_{jk}| \ge \sqrt{n}/\overline{R}] - {{\,\mathrm{\mathbb {E}}\,}}{{\,\mathrm{\mathbb {1}}\,}}[|X_{jk}| \ge \sqrt{n}/\overline{R}]] \right) ^p \\&\qquad \le C^p \left( \left( \frac{p \overline{R}^4}{n^2} \sum _{j,k=1}^n {{\,\mathrm{\mathbb {E}}\,}}|X_{jk}|^{4} \right) ^\frac{p}{2} + \frac{p^p \overline{R}^4}{n^2} \sum _{j,k=1}^n {{\,\mathrm{\mathbb {E}}\,}}|X_{jk}|^{4} \right) \le (C \sqrt{p} \overline{R}^2)^p. \end{aligned}$$

From these inequalities we may conclude the statement of Lemma. $\square$

Lemma A.2

Assuming the conditions $\mathbf{(C0)}$ we have for all $1 \le p \le A_1 \log n$

$$\begin{aligned} {{\,\mathrm{\mathbb {E}}\,}}|\tilde{m}_n(z) - {\breve{m}}_n(z)|^p \le \frac{(C \overline{R}^2)^p {\mathcal {A}}^p(2p)}{(nv)^p}. \end{aligned}$$

Proof

It is easy to see that

$$\begin{aligned} \tilde{\mathbf{R}}(z) = (\tilde{\mathbf{W}} - z\mathbf{I})^{-1} = \sigma ^{-1} (\breve{\mathbf{W}} - z \sigma ^{-1}\mathbf{I})^{-1} = \sigma ^{-1}{\breve{\mathbf{R}}}(\sigma ^{-1}z). \end{aligned}$$

(A.1)

Applying the resolvent equality we get

$$\begin{aligned} {\breve{\mathbf{R}}}(z) - {\breve{\mathbf{R}}}(\sigma ^{-1}z) = (z - \sigma ^{-1}z) {\breve{\mathbf{R}}}(z) {\breve{\mathbf{R}}}(\sigma ^{-1}z). \end{aligned}$$

(A.2)

From (A.1) and (A.2) we may conclude

$$\begin{aligned} |\tilde{m}_n(z) - {\breve{m}}_n(z)|= & {} \frac{1}{n} | {{\,\mathrm{Tr}\,}}\tilde{\mathbf{R}}(z) - {{\,\mathrm{Tr}\,}}{\breve{\mathbf{R}}}(z)| = \frac{1}{n} | \sigma ^{-1}{{\,\mathrm{Tr}\,}}{\breve{\mathbf{R}}}(\sigma ^{-1}z) - {{\,\mathrm{Tr}\,}}{\breve{\mathbf{R}}}(z)|\\= & {} \frac{1}{n} | \sigma ^{-1}{{\,\mathrm{Tr}\,}}{\breve{\mathbf{R}}}(z) - {{\,\mathrm{Tr}\,}}{\breve{\mathbf{R}}}(z)- (z - \sigma ^{-1}z) {{\,\mathrm{Tr}\,}}{\breve{\mathbf{R}}}(z) {\breve{\mathbf{R}}}(\sigma ^{-1}z)|\\\le & {} \frac{1}{n}(\sigma ^{-1} - 1) |{{\,\mathrm{Tr}\,}}{\breve{\mathbf{R}}}(z)| + (\sigma ^{-1} - 1) \frac{|z|}{n}|{{\,\mathrm{Tr}\,}}{\breve{\mathbf{R}}}(z) {\breve{\mathbf{R}}}(\sigma ^{-1}z)|. \end{aligned}$$

Taking the $p$-th power and mathematical expectation we get

$$\begin{aligned} {{\,\mathrm{\mathbb {E}}\,}}|\tilde{m}_n(z) - {\breve{m}}_n(z)|^p\le & {} \frac{1}{n^p}(\sigma ^{-1} - 1)^p {{\,\mathrm{\mathbb {E}}\,}}|{{\,\mathrm{Tr}\,}}{\breve{\mathbf{R}}}(z)|^p\\&+ (\sigma ^{-1} - 1)^p \frac{C^p}{n^p}{{\,\mathrm{\mathbb {E}}\,}}|{{\,\mathrm{Tr}\,}}{\breve{\mathbf{R}}}(z) {\breve{\mathbf{R}}}(\sigma ^{-1}z)|^p. \end{aligned}$$

Since ${\breve{\mathbf{X}}}$ satisfies conditions $\mathbf{(C1)}$ we may apply Lemma 3.1 and conclude

$$\begin{aligned} \frac{1}{n^p} {{\,\mathrm{\mathbb {E}}\,}}|{{\,\mathrm{Tr}\,}}{\breve{\mathbf{R}}}(z)|^p \le C_0^p. \end{aligned}$$

We also have

$$\begin{aligned} \sigma ^{-1} - 1 \le \sigma ^{-1} (1 - \sigma )\le & {} \sigma ^{-1} (1 - \sigma ^2) \le \sigma ^{-1} {{\,\mathrm{\mathbb {E}}\,}}|X_{jk}|^2 {{\,\mathrm{\mathbb {1}}\,}}[|X_{jk}| \ge \sqrt{n}/\overline{R}] \nonumber \\\le & {} C \overline{R}^2/n. \end{aligned}$$

(A.3)

To finish the proof it remains to estimate the term

$$\begin{aligned} \frac{1}{n^p}{{\,\mathrm{\mathbb {E}}\,}}|{{\,\mathrm{Tr}\,}}{\breve{\mathbf{R}}}(z) {\breve{\mathbf{R}}}(\sigma ^{-1}z)|^p. \end{aligned}$$

Applying the obvious inequality $|{{\,\mathrm{Tr}\,}}\mathbf{A}\mathbf{B}| \le \Vert \mathbf{A}\Vert _2 \Vert \mathbf{B}\Vert _2$ we get

$$\begin{aligned} \frac{1}{n^p}{{\,\mathrm{\mathbb {E}}\,}}|{{\,\mathrm{Tr}\,}}{\breve{\mathbf{R}}}(z) {\breve{\mathbf{R}}}(\sigma ^{-1}z)|^p\le & {} \frac{1}{n^p} {{\,\mathrm{\mathbb {E}}\,}}^\frac{1}{2}\Vert {\breve{\mathbf{R}}}(z)\Vert _2^{2p} {{\,\mathrm{\mathbb {E}}\,}}^\frac{1}{2}\Vert {\breve{\mathbf{R}}}(\sigma ^{-1}z) \Vert _2^{2p}\\\le & {} \frac{{{\,\mathrm{\mathbb {E}}\,}}^\frac{1}{2} {{\,\mathrm{Im}\,}}^p {\breve{m}}_n(z) {{\,\mathrm{\mathbb {E}}\,}}^\frac{1}{2} {{\,\mathrm{Im}\,}}^p {\breve{m}}_n(\sigma ^{-1} z)}{v^p}. \end{aligned}$$

From this inequality and (A.3) we conclude the statement of the lemma. $\square$

Lemma A.3

Assuming the conditions $\mathbf{(C0)}$ we have for all $1 \le p \le A_1 \log n$:

$$\begin{aligned} {{\,\mathrm{\mathbb {E}}\,}}|\tilde{m}_n(z) - \hat{m}_n(z)|^p \le \frac{(C\overline{R}^3)^p}{(nv)^{3p/2}}. \end{aligned}$$

Proof

It is easy to see that

$$\begin{aligned} \tilde{m}_n(z) - \hat{m}_n(z) = \frac{1}{n} {{\,\mathrm{Tr}\,}}(\tilde{\mathbf{W}} - \hat{\mathbf{W}}) \hat{\mathbf{R}} \tilde{\mathbf{R}}. \end{aligned}$$

Applying the obvious inequalities $|{{\,\mathrm{Tr}\,}}\mathbf{A}\mathbf{B}| \le \Vert \mathbf{A}\Vert _2 \Vert \mathbf{B}\Vert _2$ and $\Vert \mathbf{A}\mathbf{B}\Vert _2 \le \Vert \mathbf{A}\Vert \Vert \mathbf{B}\Vert _2$ we get

$$\begin{aligned} |\tilde{m}_n(z) - \hat{m}_n(z)| \le \Vert \tilde{\mathbf{W}} - \hat{\mathbf{W}}\Vert _2 \Vert \hat{\mathbf{R}}\Vert _2 \Vert \tilde{\mathbf{R}}\Vert = \Vert {{\,\mathrm{\mathbb {E}}\,}}\hat{\mathbf{W}}\Vert _2 \Vert \tilde{\mathbf{R}}\Vert _2 \Vert \hat{\mathbf{R}}\Vert . \end{aligned}$$

From

$$\begin{aligned} |{{\,\mathrm{\mathbb {E}}\,}}\hat{X}_{jk}| = |{{\,\mathrm{\mathbb {E}}\,}}X_{jk} {{\,\mathrm{\mathbb {1}}\,}}[|X_{jk}| \ge \sqrt{n}/\overline{R}]| \le \frac{C \overline{R}^3}{n^{3/2}} \end{aligned}$$

we obtain

$$\begin{aligned} \Vert {{\,\mathrm{\mathbb {E}}\,}}\hat{\mathbf{W}}\Vert _2 \le \frac{C \overline{R}^3}{n}. \end{aligned}$$

By Lemma A.2, we know ${{\,\mathrm{\mathbb {E}}\,}}|\tilde{m}_n(z)|^p \le C^p$. This implies that

$$\begin{aligned} \frac{1}{n^\frac{p}{2}} {{\,\mathrm{\mathbb {E}}\,}}\Vert \tilde{\mathbf{R}}\Vert _2^p \le \frac{C^p}{v^\frac{p}{2}}. \end{aligned}$$

Finally

$$\begin{aligned} {{\,\mathrm{\mathbb {E}}\,}}|\tilde{m}_n(z) - \hat{m}_n(z)|^p \le \frac{C^p \overline{R}^{3p} }{(nv)^\frac{3p}{2}}. \end{aligned}$$

$\square$

1.2 Replacement

We say that the conditions $\mathbf{(CG)}$ are satisfied if $X_{jk}$ satisfies the conditions $\mathbf{(C0)}$ and have a sub-Gaussian distribution. It is well-known that the random variables $\xi$ are sub-Gaussian if and only if ${{\,\mathrm{\mathbb {E}}\,}}^{1/p} |\xi |^p \le C \sqrt{p}$ for some constant $C>0$.

Lemma A.4

For all $v \ge v_0$ and $5 \le p \le \log n$, there exist positive constants $C_1, C_2$ such that

$$\begin{aligned} {{\,\mathrm{\mathbb {E}}\,}}|\mathbf{G}_{jk}(v)|^p \le C_1^p + C_2{{\,\mathrm{\mathbb {E}}\,}}|\mathbf{G}_{jk}^\mathbf{y}(v)|^p, \end{aligned}$$

where $G_{jk}^\mathbf{y}$ is defined in (3.26).

Proof

The method is based on the following replacement scheme, which has been used in recent results [5, 17, 20]. We replace all $h_{ab}$ by $\overline{h}_{ab}$ for $(a,b)$ such that $L_{ab} = 1$, thus replacing the corresponding resolvent entries $\mathbf{G}_{jk}$ by $\mathbf{G}_{jk}^\mathbf{y}$ for every pair of $(j, k)$. Let $\mathbb {J}, \mathbb {K}\subset \mathbb {T}$. Denote by $\mathbf{H}^{(\mathbb {J}, \mathbb {K})}$ the random matrix $\mathbf{H}$ with all entries in the positions $(\mu , \nu ), \mu \in \mathbb {J}, \nu \in \mathbb {K}$ replaced by $\overline{\xi }_{\mu \nu }$. Assume that we have already exchanged all entries in positions $(\mu , \nu ), \mu \in \mathbb {J}, \nu \in \mathbb {K}$ and are going to replace an additional entry in the position $(a, b), a \in \mathbb {T}\setminus \mathbb {J}, b \in \mathbb {T}\setminus \mathbb {K}$ with $L_{ab} = 1$. Without loss of generality, we may assume that $\mathbb {J}= \emptyset , \mathbb {K}= \emptyset$ (hence $\mathbf{H}^{(\mathbb {J}, \mathbb {K})} = \mathbf{H}$) and then denote $\mathbf{V}: = \mathbf{H}^{(\{a\}, \{b\})}$. The following additional notations will be needed.

$$\begin{aligned} \mathbf{E}^{(a,b)} = {\left\{ \begin{array}{ll} \mathbf{e}_{a} \mathbf{e}_b^{\textsf {T}} + \mathbf{e}_b \mathbf{e}_a^{\textsf {T}}, &{}\quad 1 \le a < b \le n, \\ \mathbf{e}_{a} \mathbf{e}_a^{\textsf {T}}, &{}\quad a = b. \end{array}\right. } \end{aligned}$$

and $\mathbf{U}: = \mathbf{H}- \mathbf{E}^{(a,b)}$, where $\mathbf{e}_j$ denotes a unit column-vector with all zeros except $j$-th position. In these notations, we may write

$$\begin{aligned} \mathbf{H}= \mathbf{U}+ \xi _{ab} \mathbf{E}^{(a,b)}, \quad \mathbf{V}= \mathbf{U}+ \overline{\xi }_{ab} \mathbf{E}^{(a,b)}. \end{aligned}$$

Recall that $\mathbf{G}: = (n^{-1/2}\mathbf{H}- z \mathbf{I})^{-1}$ and denote $\mathbf{S}: = (\mathbf{V}- z \mathbf{I})^{-1}$ and $\mathbf{T}: = (\mathbf{U}- z \mathbf{I})^{-1}$. Let us assume that we have already proved the following fact

$$\begin{aligned} {{\,\mathrm{\mathbb {E}}\,}}|\mathbf{G}_{jk}|^p = {\mathcal {I}}(p) + \frac{\theta _1 C^p}{n^2} + \frac{\theta _1{{\,\mathrm{\mathbb {E}}\,}}|\mathbf{G}_{jk}|^p}{n^2}, \end{aligned}$$

(A.4)

where ${\mathcal {I}}(p)$ is some quantity depending on $p, n$ (see (A.9) for precise definition) and $|\theta _1|\le 1, C > 0$ are some numbers. Similarly,

$$\begin{aligned} {{\,\mathrm{\mathbb {E}}\,}}|\mathbf{S}_{jk}|^p = {\mathcal {I}}(p) + \frac{\theta _2 C^p}{n^2} + \frac{\theta _2{{\,\mathrm{\mathbb {E}}\,}}|\mathbf{S}_{jk}|^p}{n^2}, \end{aligned}$$

(A.5)

where $|\theta _2| \le 1$. It follows from (A.4) and (A.5) that

$$\begin{aligned} \left( 1 - \frac{\theta _1}{n^2}\right) {{\,\mathrm{\mathbb {E}}\,}}|\mathbf{G}_{jk}|^p \le \left( 1 - \frac{\theta _2}{n^2}\right) {{\,\mathrm{\mathbb {E}}\,}}|\mathbf{S}_{jk}|^p + \frac{2C^p}{n^2}. \end{aligned}$$

Let us denote $\rho : = \left( 1 - \theta _2/n^2 \right) \left( 1 -\theta _1/n^2\right) ^{-1}$. We get

$$\begin{aligned} {{\,\mathrm{\mathbb {E}}\,}}|\mathbf{G}_{jk}|^p \le \rho {{\,\mathrm{\mathbb {E}}\,}}|\mathbf{S}_{jk}|^p + C_1^p/n^2, \end{aligned}$$

(A.6)

with some positive constant $C_1$. Repeating (A.6) recursively for $(a,b): L_{ab} = 1$, we arrive at the following bound

$$\begin{aligned} {{\,\mathrm{\mathbb {E}}\,}}|\mathbf{G}_{jk}|^p \le \rho ^\frac{n(n+1)}{2} {{\,\mathrm{\mathbb {E}}\,}}|\mathbf{G}_{jk}^\mathbf{y}|^p + \frac{C_1^p}{n^2} \left( 1 + \rho _1 + \cdots + \rho _1^{M - 1 }\right) , \end{aligned}$$

(A.7)

where $M \le n(n+1)/2$. It is easy to see from the definition of $\rho$ that for some $\theta$, say $|\theta | < 4$, we have

$$\begin{aligned} \rho \le 1 + |\theta |/n^2. \end{aligned}$$

From this inequality and (A.7), we deduce that

$$\begin{aligned} {{\,\mathrm{\mathbb {E}}\,}}|\mathbf{G}_{jk}|^p \le C_2 {{\,\mathrm{\mathbb {E}}\,}}|\mathbf{G}_{jk}^\mathbf{y}|^p + C_3^p, \end{aligned}$$

with some positive constants $C_2$ and $C_3$. From the last inequality, we may conclude the statement of the lemma. It remains to prove (A.4) (resp. (A.5)). Applying the resolvent equation, we get for $m \ge 0$

$$\begin{aligned} \mathbf{G}= \mathbf{T}+ \sum _{\mu =1}^m \frac{(-1)^\mu }{n^\frac{\mu }{2}} \xi _{ab}^\mu (\mathbf{T}\mathbf{E}^{(a,b)})^\mu \mathbf{T}+ \frac{(-1)^{m+1}}{n^\frac{m+1}{2}} \xi _{ab}^{m+1} (\mathbf{T}\mathbf{E}^{(a,b)})^{m+1} \mathbf{G}. \end{aligned}$$

(A.8)

The same identity holds for $\mathbf{S}$

$$\begin{aligned} \mathbf{S}= \mathbf{T}+ \sum _{\mu =1}^m \frac{(-1)^\mu }{n^\frac{\mu }{2}} \overline{\xi }_{ab}^\mu (\mathbf{T}\mathbf{E}^{(a,b)})^\mu \mathbf{T}+ \frac{(-1)^{m+1}}{n^\frac{m+1}{2}} \overline{\xi }_{ab}^{m+1} (\mathbf{T}\mathbf{E}^{(a,b)})^{m+1} \mathbf{S}. \end{aligned}$$

We investigate (A.8). In order handle arbitrary high moments of $\mathbf{G}_{jk}$, we apply a Stein type technique similar to Theorem. Let us introduce the following function $\varphi (z): = \overline{z} |z|^{p-2}$ and write

$$\begin{aligned} {{\,\mathrm{\mathbb {E}}\,}}|\mathbf{G}_{jk}|^p = {{\,\mathrm{\mathbb {E}}\,}}\mathbf{G}_{jk} \varphi (\mathbf{G}_{jk}). \end{aligned}$$

Applying (A.8), we get

$$\begin{aligned} {{\,\mathrm{\mathbb {E}}\,}}|\mathbf{R}_{jk}|^p= & {} \sum _{\mu =0}^4 \frac{(-1)^\mu }{n^\frac{\mu }{2}} {{\,\mathrm{\mathbb {E}}\,}}\xi _{ab}^\mu [(\mathbf{T}\mathbf{E}^{(a,b)})^\mu \mathbf{T}]_{jk} \varphi (\mathbf{G}_{jk}) \nonumber \\&+\sum _{\mu =5}^m \frac{(-1)^\mu }{n^\frac{\mu }{2}} {{\,\mathrm{\mathbb {E}}\,}}\xi _{ab}^\mu [(\mathbf{T}\mathbf{E}^{(a,b)})^\mu \mathbf{T}]_{jk} \varphi (\mathbf{G}_{jk}) \nonumber \\&+ \frac{1}{n^\frac{m+1}{2}} {{\,\mathrm{\mathbb {E}}\,}}\xi _{ab}^{m+1} [(\mathbf{T}\mathbf{E}^{(a,b)})^{m+1} \mathbf{G}]_{jk}\varphi (\mathbf{G}_{jk}) =: {\mathcal {A}}_0 + {\mathcal {A}}_1 + {\mathcal {A}}_2. \end{aligned}$$

Repeating the arguments from [17], one may show that

$$\begin{aligned} \max (|{\mathcal {A}}_1|, |{\mathcal {A}}_2|) \le \frac{C^p}{n^2} + \rho \frac{{{\,\mathrm{\mathbb {E}}\,}}|\mathbf{G}_{jk}|^p}{n^2}. \end{aligned}$$

For the term ${\mathcal {A}}_{00}$, one may write down the following representation

$$\begin{aligned} {\mathcal {A}}_0 = {\mathcal {I}}(p) + r_n(p), \end{aligned}$$

with the remainder term bounded in absolute value

$$\begin{aligned} |r_n(p)| \le \frac{C^p}{n^2} + \rho \frac{{{\,\mathrm{\mathbb {E}}\,}}|\mathbf{G}_{jk}|^p}{n^2}, \end{aligned}$$

and

$$\begin{aligned} {\mathcal {I}}(p) := \sum _{\mu =0}^4 \frac{(-1)^\mu }{n^\frac{\mu }{2}} {{\,\mathrm{\mathbb {E}}\,}}\xi _{ab}^\mu {{\,\mathrm{\mathbb {E}}\,}}[(\mathbf{T}\mathbf{E}^{(a,b)})^\mu \mathbf{T}]_{jk} \varphi (\mathbf{T}_{jk}) + \sum _{\mu =0}^4 \sum _{l=1}^{4-\mu } \frac{(-1)^\mu }{l!} {\mathcal {B}}_{\mu l}^{(0)},\nonumber \\ \end{aligned}$$

(A.9)

where

$$\begin{aligned} {\mathcal {B}}_{\mu l}^{(0)}:= & {} \sum _{\begin{array}{c} \mu _1 + \cdots + \mu _m = l \\ \mu +\mu _1 + 2\mu _2 + \cdots + m \mu _m \le 4 \end{array}} \frac{C_{\mu _1, \ldots , \mu _m}^l}{n^{\frac{\mu }{2} + \frac{\mu _1}{2} + \frac{2\mu _2}{2} + \cdots + \frac{m \mu _m}{2}}} {{\,\mathrm{\mathbb {E}}\,}}X_{ab}^{\mu +\mu _1 + 2\mu _2 + \cdots + m \mu _m} \\&\qquad \qquad \times {{\,\mathrm{\mathbb {E}}\,}}[(\mathbf{T}\mathbf{E}^{(a,b)})^\mu \mathbf{T}]_{jk}[(\mathbf{T}\mathbf{E}^{(a,b)})\mathbf{T}]_{jk}^{\mu _1}\ldots [(\mathbf{T}\mathbf{E}^{(a,b)})^m\mathbf{T}]_{jk}^{\mu _m} \varphi ^{(l)}(\mathbf{T}_{jk}). \end{aligned}$$

One may see that the term ${\mathcal {I}}(p)$ does not depend on $\mathbf{G}$ but depends on $\mathbf{T}$. $\square$

Lemma A.5

Let $\mathbf{L}$ be $r$-admissible and assume that the conditions $\mathbf{(CG)}$ hold. Let $C_0$ and $s_0$ be arbitrary numbers such that $C_0 \ge \max (1/V, 6 c_0), s_0 \ge 2$. There exist a sufficiently large constant $A_0$ and small constant $A_1$ depending on $C_0, s_0, V$ only such that the following statement holds. Fix some $\tilde{v}: \tilde{v}_0 s_0 \le \tilde{v} \le V$. Suppose that for some integer $L > 0$, all $u, v',q$ such that $\tilde{v} \le v' \le V,\, |u| \le u_0, 1 \le q \le A_1 (n v')$

$$\begin{aligned} \max _{\mathbb {J}: |\mathbb {J}| \le L} \max _{l, k \in \mathbb {T}_\mathbb {J}}{{\,\mathrm{\mathbb {E}}\,}}|\mathbf{G}_{l k}^{(\mathbb {J})}(v')|^q \le C_0^q. \end{aligned}$$

Then for all $u,v, q$ such that $\tilde{v}/s_0 \le v \le V, |u| \le u_0$, $1 \le q \le A_1 (n v)$

$$\begin{aligned} \max _{\mathbb {J}: |\mathbb {J}| \le L-1} \max _{l, k \in \mathbb {T}_\mathbb {J}}{{\,\mathrm{\mathbb {E}}\,}}|\mathbf{G}_{l k}^{(\mathbb {J})}(v)|^q \le C_0^q. \end{aligned}$$

Proof

We first observe the fact that the factor $q$ appears only in the terms with $\overline{\xi }_{jk}$. Let us consider only one term, for example, :

$$\begin{aligned} \zeta _{2j}:= -\frac{2}{n}\sum _{l \ne k \in \mathbb {T}_{\mathbb {J},j} \setminus \mathbb {A}_j} (\overline{\xi }_{jl} - {{\,\mathrm{\mathbb {E}}\,}}\overline{\xi }_{jl}) (\overline{\xi }_{jk} - {{\,\mathrm{\mathbb {E}}\,}}\overline{\xi }_{jk}) \mathbf{G}_{kl}^{(\mathbb {J}, j)}. \end{aligned}$$

Applying the Hanson–Wright inequality, see, e.g., [22] we obtain that

$$\begin{aligned} {{\,\mathrm{\mathbb {E}}\,}}|\zeta _{2j}|^{2q} \le \frac{C^q q^q {{\,\mathrm{\mathbb {E}}\,}}{{\,\mathrm{Im}\,}}^q m_n^{(\mathbb {J},j)}}{(nv)^q} + \frac{C^q q^{2q}}{(nv)^{2q}} \le \frac{C^q q^q (C_0 s_0)^q }{(nv)^q}. \end{aligned}$$

$\square$

1.3 Inequalities for Resolvent

Lemma A.6

For any $z = u + i v \in \mathbb {C}^{+}$, we have

$$\begin{aligned} \frac{1}{n} \sum _{l,k \in \mathbb {T}_{\mathbb {J}}} |\mathbf{R}_{kl}^{(\mathbb {J})}|^2 \le \frac{1}{v} {{\,\mathrm{Im}\,}}m_n^{(\mathbb {J})}(z). \end{aligned}$$

(A.10)

For any $l \in \mathbb {T}_{\mathbb {J}}$

$$\begin{aligned} \sum _{k \in \mathbb {T}_{\mathbb {J}}} |\mathbf{R}_{kl}^{(\mathbb {J})}|^2 \le \frac{1}{v} {{\,\mathrm{Im}\,}}\mathbf{R}_{ll}^{(\mathbb {J})}. \end{aligned}$$

(A.11)